Class 12 Chemistry Haloalkanes and Haloarenes NCERT Solutions – Download PDF

(i) Bromocyclohexane is reacting with the metal Mg to give organo-metallic compound/Grignard reagent

i.e. Cyclohexyl Magnesium Bromide (A).

Grignard reagents are highly reactive with hydrocarbons to give alkanes. Cyclohexyl Magnesium Bromide reacts with water to cyclohexane (B) and Mg (OH)Br.

(ii) As the above reaction, haloalkane with metal Mg react to form the Grignard reagent, alkyl magnesium halide, RMgX, and further with hydrocarbon gives alkane.

∴ C is CH₃CH (MgBr)CH₃, a Grignard Reagent, R is CH₃CH (Br)CH₃ and D ≅ H

(iii) In the Wurtz Reaction, alkyl halide in the presence of sodium in dry ether gives the double number of

Now, in the reaction 2,2,3,3-tetramethyl butane is formed after the alkyl halide reacting in the presence of sodium in dry ether.

∴ Number of carbons will be 3 in R¹ –X is CH₃C (X) (CH₃)CH₃

And the remaining reaction is same as the above i.e. alkyl halide reacting with Magnesium to give Grignard Reagent (D) and further with hydrocarbon H₂O to give alkane (E)

∴ D is CH₃C (CH₃) (CH₃)MgBr E is CH₃CH (CH₃)CH₃

(i) Since I^- ion is a better leaving group than Br^- ion, therefore, CH₃I reacts faster CH₃Br in S_N2 reaction with OH^- ion.

Better the leaving group, faster is the S_N2 reaction.

(ii) On steric grounds, 1^? alkyl halides are more reactive than tert-alkyl halides in S_N2 reactions. Therefore, CH₃Cl will react at a faster rate than (CH₃)₃CCl in a S_N2 reaction with OH- ion. (Bulkier the group, slower is the S_N2 )

(a) 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final

(b) The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.

(c) The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1- iodobutane as the final product and NaBr as the by-product.

The reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered (does not have any free space or very crowded), then the reactivity towards S_N2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

• Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane. The structures are shown below:

Hence, the increasing order of reactivity towards S_N2 displacement is:

• Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
• The stearic hinderance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is given as-

3° < 2° < 1°.

The structures are given below:

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards S_N2 displacement as:

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

• The stearic hinderance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the stearic hinderance increases with an increase in the number of Therefore, the increasing order of steric hindrances in the given compounds is as below:

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1- Bromobutane

(i) Propene to propan-1-ol

It is an antimarkovnikoff reaction in which under the presence of a peroxide an alkene undergoes substitution, wherein the halo group is attached to that carbon which has least no. Of alkyl groups attached to it.

There are two primary alkyl halides having the formulaC₄H₉Br, They are n - butyl bromide and isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C₁₈H₁₈ which is different from the compound formed when n-butyl bromide reacts with Na metal.

Hence, compound (a) must be isobutyl bromide. Thus, compound (d) is 2, 5-dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2- methylpropene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2-bromo-2-methylpropane.

Dipole moment of a molecule depends on the electronegativity difference between atoms bonded covalently and geometry of the molecule (how far the atoms are from each other). Dipole moment is important to understand the polarity of a molecule.

The three dimensional structures of the three compounds along with the direction of dipole moment in each of their bonds are given below:-

CCl₄ being symmetrical has zero dipole moment. In CHCl₃, the resultant of the two C-Cl dipole moments is opposed by the resultant of C-H and C-Cl bonds. Since the dipole Moment of latter resultant is expected to be smaller than the former, CHCl₃ has a finite dipole moment (1.03D)

In CH₂Cl_2, the resultant of two C-Cl dipole moments is reinforced by resultant of two C-H dipoles. Therefore, CH₂Cl₂ (1.62D) has a dipole moment higher than that of CHCl₃. Thus, CH₂Cl₂ has the highest dipole moment.

The given reaction is a nucleophillic reaction:

The given reaction is an SN2 reaction. In this reaction, CN acts as the stronger nucleophile and attacks the carbon atom to which Br is attached in nBuBr.

And, CN^- ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

The mechanism is shown below:

10.3 There are four different dihalogen derivatives of propane. The structures of these derivatives are as shown below: -

This is a classic question from the chapter Haloalkanes and Haloarenes. Let us solve each one as follows:

(1) When n - butyl chloride is treated with alcoholic KOH, the formation of but - l - ene takes place. This reaction is a dehydrohalogenation reaction.

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-\mathrm{Cl}\underset{\text{KOH (alc), Δ}}{\overset{\text{Dehydrohalogenation}}{\to }}C{H}_3-C{H}_2-CH=C{H}_2+\mathrm{KCl}+H_{2}O$

When N-butyl chloride or 1-chlorobutane is treated with alcoholic potassium hydroxide (KOH), elimination reaction takes place. This leads to alkene formation. 1-butene is the major product here. 

(ii) When Bromobenzene is treated with Mg in the presence of dry ether, it undergoes a reaction to produce phenylmagnesium bromide which is the Grignard reagent. This reagent is very reactive organomagnesium compound. Ether solvent is important since it produces a complex with Grignard reagent.

(iii) Chlorobenzene is aryl halide does not easily react with dilute adequate bases or water. However, under extreme conditions like high temperature and pressure and in the presence of strong base/acid catalyst, chlorobenzene undergoes hydrolyosis to form phenol.

The Haloalkenes and Haloarenes NCERT excercise covers many different types of reaction based questions which have been even asked in competitive exams like JEE Main entrance exam.

 

(iv) When Ethyl Chloride or Chloroethane is treated with the aqueous KOH, a nucleophilic substitution reaction (mostly $S_{N}2$ ) take place. This leads to the formation of ethanol.

 

$C{H}_{3}C{H}_2\mathrm{Cl}+KOH\text{\hspace{0.17em}(aq)}\to C{H}_{3}C{H}_{2}OH+K\mathrm{Cl}$

CH₃CH (Cl)CH (Br)CH₃: 2-Bromo-3-chlorobutane

1. Write the name of side substituent according to the alphabetical order. Example: bromo is written before chloro as B comes before
2. Always use a hyphen {-} between a number and a Since Br is attached to 2 position and Cl is attached to third position i.e. it is written as 2-Bromo and 3-chloro.
3. Lastly, write the name of the longest hydrocarbon chain e. butane of 4 carbons.

CHF₂CBrClF :1-bromo-1-chloro-1,2,2-trifluoroethane

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical Eg: bromo is written before chloro and fluoro as B comes before C and F.
2. Always use a hyphen {-} between a number and a Since Br is attached to first carbon and Cl Is also attached to first carbon i.e. it is written as 1-Bromo and 1-chloro.
3. Since there are 3 F present so it is written as
4. And at the last write the name of the longest hydrocarbon chain e. ethane of 2 carbons.

ClCHC≡CCH₂Br: 1-bromo-4-chlorobut-2-yne

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical That is why bromo is written before chloro as B comes before C.
2. Always use a hyphen {-} between a number and a Since Br is attached to 1 position and Cl isattached to fourthposition i.e. it is written as 1-Bromo and 4-chloro.
3. Also there is a presence of triple bond at the second carbon so the suffix used for triple bond is “yne” along with the position of the bond i.e. but-2-yne

(CCl₃)₃CCl

2- (trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 3 CCl₃ group tri is used as a
4. Since 7 Cl atoms are present in all therefore hepta is used a prefix with the position of chlorine. And at the last write the name of the longest hydrocarbon chain e. propane)

CH₃C (p-ClC₆H₄)₂CH (Br)CH₃

2-bromo-3,3-bis- (4-chlorophenyl)butane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 (p-ClC₆H₂) group bis is used as a prefix instead of di because C₆H₄ contains itself contains numbers 6 &
4. And at the last write the name of the longest hydrocarbon chain e. butane)

(CH₃)₃CCH=CClC₆H₄I-p

1-chloro-1- (4-iodophenyl)-3,3-dimethylbut-1-ene

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 methyl group di is used as a
4. And the presence of double bond has the ending ‘ene’ and the presence of double bond at the first position is written as but-1-ene.)

Let us explain each question from Haloalkane and Haloarenes chapter one by one. Students who are currently in school and plan to take the CBSE board exam for class 12th soon, need to prepare all these questions. Let us get started.

(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Sp² hybrid carbon (s-character=33.33%) in chlorobenzene is more electronegative than a sp³-hybrid carbon (s-character=25%) in cyclohexylchloride, due to greater s-character. Thus, Carbon atom of chlorobenzene has less tendency to release electrons to Cl than carbon atom of cyclohexylchloride.

Chlorine atom in chlorobenzene is attached to the benzene ring. Here, the carbon-chlorine bond is polar. However, electron delocalization is possible in the benzene ring. Due to the electron delocalization, the overall dipole moment reduced since the polarity is distributed over the entire ring structure.

On the other hand, the chlorine atom in cyclohexyl chloride is attached to a saturated carbon atom in the cyclohexane ring. Because of this, there is no electron delocalization in the saturated structure. Hence, the dipole moment is more localized and stronger as compared to chlorobenzene.

 (ii) Alkyl halides, though polar, are immiscible with water.

Alkyl halides are polar because of the difference in electronegativity between carbon and halogen atoms. Here, there is not enough polarity to make them soluble in water.

The alkyl part of molecule is non-polar as well as hydrophobic in nature. This portion is large enough to dominate the solubility characteristics of the molecule. The entire molecule becomes insoluble in water due to this.

Water molecules create extensive hydrogen bonds with each other. However, Alkyl halides cannot form hydrogen bonds with water, which reduces their solubility in water. Haloalkane and Haloarene NCERT excercise discusses this question in further detail.

(iii) Grignard reagents should be prepared under anhydrous conditions.

Grignard reagents are very reactive organometallic compounds and they react readily with water to produce hydrocarbons. This destroys the Grignard reagent and makes it ineffective for its original purpose.

The reaction with water not only destroy Grignard reagent but also it produces by products that can complicate the synthetic pathway.

Preparing Grignard reagents under anhydrous conditions makes sure that they remain stable and effective for use in further reactions. Moisture-free conditions are important for the formation and utilization of reagents.

R-Mg-X + H-OH → R-H + Mg (OH)X

Do note that questions based on such equations from organic chemistry are also be asked in the CUET exam which makes it important for students to remember it.

Thus, Grignard reagents must be prepared under anhydrous conditions.

Carbon due to its chemical property of catenation, tetravalency and the ability to form stable covalent bonds form hydrocarbon and halo-alkanes easily.

The process of removal of hydrogen and halogen from adjacent carbon atoms of haloalkens is called dehydrohalogenation. NaOEt ( Sodium ethoxide) used in Dehydrohalogenation.  It is a strong base and helps remove a β-hydrogen from the haloalkane.

The chemical reaction for

Reaction:

R–CH2 –CHX–R′+NaOEt   ethanol > R–CH=CHR’+NaX+EtOH

(i) 1-Bromo-1-methylcyclohexane: 

      Br
       |
 CH3–C1–cyclohexane  + NaOEt/EtOH → CH3–C=cyclohexane (double bond at C1-C2) + NaBr + EtOH

Major Product: 1-Methylcyclohexene

(ii) 2-Chloro-2-methylbutane

      CH3
           |
 CH3–C–CH2–CH3   + NaOEt/EtOH → CH3–C=CH–CH3 + NaCl + EtOH
           |                                 ||
          Cl                             CH3

Major Product: 2-Methyl-2-butene (major), 2-Methyl-1-butene (minor)

(iii) 2,2,3-Trimethyl-3-bromopentane

       CH3         CH3
         |            |
 CH3–C–C–CH2–CH3  + NaOEt/EtOH → CH3–C=C–CH2–CH3 + NaBr + EtOH
         |   |                            ||
        CH3  Br                        CH3

Major Product: 2,2,3-Trimethyl-2-pentene

Since ethanol is a alcohol, so in presence of alcoholic KOH, alkyl chloride undergo elimination reaction, which results in the removal of proton and being substituted by a bromide ion.

It is also a simple substitution reaction in which under the presence of aqueous NaOH, bromide ion is replaced by hydroxide ion as it is a better leaving group than hydroxide ion.

 

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.

CH₃CH₂Br + KCN - CH₃CH₂CN + kBr

5-C₆H₅ONa + C₂H5Cl

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.

C₆H₅ONa + C₂H₅Cl? C₆H₅ – O – C₂H₅ + NaCl

6-CH₃CH₂CH₂OH + SOCl₂

This reaction is called as wurtz reaction in which in which clubbing together of a alkyl halide group with a sodium or potassium salt of an ether group.

 

7-CH₃CH₂CH = CH₂+ HBr

It is an antimarkovnikoff reaction in which under presence of a peroxide an alkene undergoes substitution, wherein the halo group is attached to that carbon which has least no. Of alkyl groups attached to it.

 

8-CH₃CH = C (CH₃)₂ + HBr

It is a markovnikoff reaction in which an alkene undergoes substitution reaction with hydrohalide, wherein a halo group is attached to that carbon atom which has the largest no. Of alkyl groups attached to it.

(i) Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps.

1. In the first step ethanol is treated with thionyl chloride (SOCl₂) in the presence of pyridine to give chloroethane

• In the second step the acetylene is treated with sodamide(NaNH₂) in the presence of liquid ammonia to give sodium acetylide

• The last step involves the reaction between chloroethane and sodium acetylide to give the final product as the But-1-yne and NaCl as the by-product.

• The bromination of ethane (Br₂ in the presence of light at 520-670K) gives bromoethane which on further treatment with alcoholic KOH results in the alkene called ethane which again on bromination with Br₂/CCl₄ gives dibromide(vicinal bromide) called 1,2-dibromoethane which on treatment with alcoholic KOH gives bromoethene as the final product.

• Propene on treatment with HBr in the presence of peroxide (anti-markovnikoff’s rule which states the negative part e. Br^- goes to the carbon which has more number of hydrogens and positive part i.e. H⁺ goes to the carbon which has lesser number of hydrogens) gives 1-bromopropane which on treatment with silver nitrite in the presence of alcohol or water gives 1-nitropropane as the final product.

• Chlorination of toluene (Cl₂ at 773K) gives benzyl chloride with the removal of HCl. Benzyl Chloride on further treatment with dilute KOH in the presence of heat gives benzyl alcohol as the final (dilute KOH removes Cl from the benzyl chloride and replaces it by OH)

• Propene is treated with Br₂/CCl₄e. bromination of alkene takes place to give dibromides, in this case 1,2-dibromopropane which on further treatment with alcoholic KOH gives propyne.(two times removal of KBr)

• Ethanol in treatment with thionyl chloride(SOCl₂) in the presence of pyridine gives ethyl chloride with the evolution of SO₂ The ethyl chloride on treatment with mercury fluoride gives ethyl fluoride (the Cl of ethyl chloride is being replaced by F from Hg₂F₂).

• Bromoethane on treatment with alcoholic KCN forms Acetonitrile with the removal of KBr. CN is added in the case where there is a need to increase the number of Further the acetonitrile is treated with CH₃MgBr(Grignard reagent) in the presence of ether, the intermediate so formed is hydrolysed, and the product formation takes place i.e. Propanone and NH₃ is obtained as the by- product.

• But-1-ene reacts with HBr to form 2-bromobutane(acc. To markovnikoff’s rule) and further 2- bromobutane is treated with alcoholic KOH in the presence of heat to form But-2-ene with the removal of HBr.(alcoholic KOH removes H from the carbon which has less number of hydrogens just next to the carbon to which halide is attached and gives the major product e. Saytzeff rule)

• The conversion of 1-chlorobutane to n-octane in the presence of Na metal and dry ether is called Wurtz Two equivalent Chlorobutane reacts to give n-octane as a final product and NaCl as the by-product.

 

The conversion of benzene to biphenyl using Na and dry ether is called FITTIG reaction.

Firstly, the benzene is treated with Br₂/FeBr₃ results in the formation of bromobenzene. It is an electrophilic substitution. Further the bromobenzene is treated with Na metal in presence of dry ether to form biphenyl as the final product and NaBr as the by-product.

 

If instead of Na metal, bromobenzene is treated with Cu metal then the reaction is called Ullmann’s reaction. The final product obtained in this reaction will also be biphenyl.

A 10.5 1. Cyclohexanol will react with thionyl chloride to form Chlorocyclohexane with the evolution of sulphur dioxide and hydrogen chloride gas as shown below :

2. 4-Ethylnitrobenzene will undergo benzylic bromination in presence of heat or light. Now, as benzylic radicals are more stable therefore benzylic hydrogen is abstracted. Hence, the reaction yields 4-(1- Bromoethyl)nitrobenzene as a product

3. 4-Hydroxymethylphenol will react with hydrochloric acid under thermal conditions to yield 4- Chloromethylphenol a shown in the reaction below:

4. 1-Methylcyclohexene will react with hydrogen iodide via markovnikov’s addition mechanism to give 1- Iodo1-methylcyclohexane as shown in the reaction below: -

5.Bromoethane will react with sodium iodide in a Finkelstein reaction manner to yield Iodoethane in presence of dry acetone and heat as shown in the reaction below : 

6. Cyclohexene reacts with bromine in presence of heat and light to undergo allylic bromination to yield 3-bromocyclohex-1-ene as the major product as shown in the reaction below : 

10.1 From the name of the compound it is clear that the parent ring is pentane and chloro and methyl groups are attached in the straight chain at 2nd and 5th position respectively. Hence, the structure is as follows:

From the name of the compound given it is clear that the parent group is hexane with 2 attachments namely chloro and ethyl groups at 1 and 4 positions respectively. Hence, the structure is as follows: -

From the name of the compound given it is clear that the parent group is heptane with tertiary butyl and iodine groups attached at 4 and 3 positions respectively. Hence, the structure is as follows: -

From the name of the compound given it is clear that the parent group is an alkene with the unsaturation present in between 2^nd and 3^rd carbon and 2 bromine atoms attached at 1^st and 4^th carbon. Hence, the structure is as follows : -

From the name of the compound given it is clear that the parent group is benzene with bromine group attached at 1^st carbon, secondary butyl group attached at 4^th carbon and a methyl group attached at 2^nd carbon. Hence, the structure is a follows : -

As per the molecular formula $C_5{H}_{10}$ , one thing can be confirmed; the hydrogen is either cycloalkane or it is an alkene. However, alkenes readily react with chlorine in even dark because of the double bond. Therefore, it is possible that the hydrocarbon is cycloalkane. 

Now, let us consider the reactivity. Cycloalkanes are the saturated hydrocarbons with no double bonds. These do not react with Chlore in dark because such a reaction needs UV light for initiating the process of free radical substitution. Haloalkane and Haloarenes NCERT solutions cover this as well as other questions in further detail.

In bright sunlight, the UV light starts free radical substitution reaction. In cyclopentane, the substitution will result in single monochloro product because all hydrogen atoms are equivalent in the molecule. Therefore, by replacing any hydrogen with chlorine atom will result again in the same product i.e. $C_5{H}_9\mathrm{Cl}.$ This confirms that the hydrocarbon is cyclopentane.

In S_N2 Reaction, the product will be formed in a single step with no intermediates. Nucleophile attack will be easier for simple halides than hindered haloalkanes. Therefore,

1. Bromobutane (1°)reacts faster than 2-Bromobutane (2°)
2. 2-Bromobutane (2°) reacts faster than 2-Bromo-2-methylpropane or tert-Butyl bromide (3°) 1-Bromo-3-methyl butane reacts faster than the 1-Bromo-2-methyl bu
3. tane as the former is less hindered with respect to leaving halide than the later which is more hindered comparatively

In the presence of sulphuric acid (H₂SO₄), KI produces HI as follows : -

2KI + H₂SO₄ → 2KHSO₄ + 2KI

Since H₂SO₄ is an oxidising agent, it oxidises HI (produced in the reaction to I₂)

2HI + H₂SO_{4 →} I₂ + SO₂ +;

As a result, the reaction between an alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidising acid such as H3PO4 is used in the reaction to get the desired product.

A few things you can remember here, while solving NCERT Solutions for Haloalkanes And Haloarenes.

• Sulphuric acid is a powerful oxidising agent. We know this because the sulphur atom in sulphuric acid is in its highest possible oxidation state (+6). It can only be reduced. And it can readily accept electrons from other substances.

• It oxidises the necessary intermediate, hydrogen iodide (HI), to iodine (I2). We know this because iodide ions are strong reducing agents, and they can simply give up electrons. So when sulphuric acid accepts electrons, it can convert the iodide to elemental iodine.

• The alcohol cannot be converted to alkyl iodide. The reason for this is that the side reaction consumes the iodide ions necessary to act as a nucleophile and replace the alcohol's -OH group. This is the desired alkyl iodide.

• A non-oxidising acid like phosphoric acid (H3PO4) is used instead to produce the desired alkyl iodide. Phosphoric acid is a weaker oxidising agent than sulphuric acid and does not react with the iodide ions. That allows the intended reaction with the alcohol to proceed.  
  
  
  

An aqueous solution, KOH almost completely ionizes to give OH-ions. OH- ion is a strong nucleophile (see the imp. note below), which leads the alkyl chloride to undergo a nucleophilic substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO-) ion, it is a strong base. Thus, it can abstract a hydrogen ion from the beta-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.OH- ion is a much weaker base than RO- ion. Also, OH- ion is highly solvated (because more energy is released on solvation)in an aqueous solution and as a result, the basic character of OH-ion decreases.

Therefore, it cannot abstract a hydrogen from the beta -carbon.

The concept used hydroxide ion is a strong nucleophile but weaker base than -OR group.

Reason: SN₁ occurs in two steps. In the first step, carbocation forms. The rate of reaction in S_N1 depends upon the stability of the carbocation, greater the stability faster the reaction. Tertiary (3°) carbocation is more stable than secondary (2°), which is further stable than primary (1°)

By using this

1. Tert-Butyl Chloride (3°) reacts faster than 3-Chloropentane (2°)
2. 2-Chloro heptane (2°) is more reactive than 1-Chloro hexane (1°)

Hydrolysis by KOH results in the formation of the carbocation. Those compounds which leads to the formation of stable carbocation are easily hydrolysed.C₆H₅CH₂Cl leads to formation of 1°- carbocation, while C₆H₅CHClC₆H₅ forms 2°-carbocation, which is more stable than 1°-carbocation. Hence C₆H₅CHClC₆H₅, is hydrolyzed more easily than C₆H₅CH₂Cl by aqueous KOH.

(i) 2-chloro-3-methylbutane, (2^? alkyl halide)

Explanation: The Cl atom is attached with two alkyl groups therefore 2^? alkyl halide.

(ii)  3-chloro-4-methylhexane (2^? alkyl halide)

Explanation: The Cl atom is attached with two alkyl groups therefore 2^? alkyl halide.

(iii)  1-iodo-2,2-dimethylbutane (1^? alkyl halide)

Explanation: The atom I is attached with one alkyl group therefore 1^? alkyl halide.

(iv)  1-bromo-3,3-dimethyl-1-phenylbutane (2^? benzylic halide)

Explanation: The atom Br is attached with two alkyl groups with the benzene group therefore 2^? benzylic halide.

(v)  2-bromo-3-methylbutane (2^? alkyl halide)

Explanation: The atom Br is attached with two alkyl groups therefore 2^? alkyl halide.

(vi)  1-bromo-2-ethyl-2-methylbutane (1^? alkyl halide)

Explanation: The atom Br is attached with one alkyl group therefore 1^? alkyl halide.)

(vii)  3-chloro-3-methylpentane (3^? alkyl halide)

Explanation: The atom Cl is attached with three alkyl groups therefore 3^? alkyl halide.

(viii)  3-chloro-5-methylhex-2-ene (vinylic halide)

Explanation: The atom Cl is attached with the double bond i.e. at vinylic position ∴ vinylic halide.)

(ix)  4-bromo-4-methylpent-2-ene (allylic halide)

Explanation: The Br atom is attached with the carbon next to the double bond i.e. at allylic position therefore allylic halide.

(x)  1-chloro-4- (2-methylpropyl)benzene (aryl halide)

Explanation: The Cl atom is attached with the benzene i.e. at aryl position therefore aryl halide.

(xi)  1-chloro-3- (2,2-dimethylpropyl)benzene (1^? benzylic halide)

Explanation: The Cl is attached with one alkyl group at the benzene carbon therefore 1^? benzylic halide.

(xii)  1-bromo-2- (1-methylpropyl)benzene (aryl halide)

Explanation: The Br is attached with the benzene i.e. at aryl position therefore aryl halide.

Double bond equivalent is used to find the level of unsaturations present in an organic molecule.

DBE = C +1-H/2+X/2-N/2

Where C= number of carbon atoms present H=number of hydrogen atoms present N=number of nitrogen atoms present X=number of halogen atoms present Double bond equivalent (DBE) for C₄H₉Br

= 4+1-9/2+1/2

=0

So none of the isomers has a ring or unsaturation, so the isomers are positions or chain isomers as shown below in the table:

Ambident Nucleophiles are those nucleophilies which can attack through different sites. For example:-cyanide ions are a resonance hybrid of the following two structures:

It can attack through carbon to form cyanide and through N to form is O cyanide. Example: NO₂, NO₃ ^- etc.

(i) Freon 12:- the chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic and non-corrosive. Example of Freon 12 is CCl₂F₂. It is mainly used in refrigeration and eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In stratosphere, Freon is able to initiate the radical chain reactions that can upset the natural ozone balance.

• DDT: - it stands for p, p'-Dichlorodiphenyltrichloroethane (DDT). It is mainly used as an
• Iodoform: - it was earlier used as an antiseptic but the antiseptic properties are due the liberation of free iodine and not due to the iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.
• Carbon tetrachloride (CCl₄):- it is used as an industrial solvent for oils, fats, resins etc. and also in dry Used as a cleaning fluid both in industries, as a degreasing agent and in the home, as a spot remover and as a fire extinguisher.

To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C₅H₁₂. This is because of the fact that replacement of any H-atom leads to the formation of the same product. Therefore the isomer is 2,2-dimethylpropane.

To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C₅H₁₂ should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane as shown below : -

To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C₅H₁₂ should contain four different types of H-atoms in it. Therefore, the required isomer is 2-methylbutane. It can be observed that there are four different types of H-atoms labelled as a, b, c, and d in 2-methylbutane as shown below: -

Here is the reasoning that you can learn more while practicing the NCERT Solutions for Haloalkanes And Haloarenes. 

 

This outcome is directly attributed to the free radical halogenation mechanism. Usually, in photochemical chlorination, a chlorine atom can replace any hydrogen atom in the alkane. The number of unique products, or isomeric monochlorides, depends on the number of distinct types of hydrogen atoms in the starting isomeric alkane. 

 

For instance, the high symmetry of 2,2-dimethylpropane (neopentane) means all its hydrogen atoms are equivalent. That itself yields only a single monochloride. But, in contrast, the different primary and secondary hydrogens in n-pentane and 2-methylbutane create multiple products.

p-Dichlorobenzene is more symmetrical than o-and m-isomers. Because it fits more closely and easily in the crystal lattice than o-and m-isomers. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. Therefore, p-dichlorobenzene is symmetrical and has a higher melting point and lower solubility than o-and m-isomers due to strong force of attraction in crystal.

Therefore more energy required to break lattice and will not easily be soluble.

The order of increasing boiling point is

1. Chloromethane (CH₃Cl)< Bromomethane (CH₃Br) < Dibromomethane (CH₂Br₂)< Bromoform (CHBr₃)
2. Isopropyl chloride (C₃H₇Cl)< 1-Chloropropane (C₃H₇Cl) < 1-Chlorobutane (C₄H₉Cl)

Generally, boiling point increases with the molecular weight of the compound and decreases with the branching of the chain